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One slit of a double slit experiment is covered by a thin glass plate of refractive index 1.4, and the other by a thin glass plate of the refractive index 1.7. The point on the screen where the central maximum fall before the glass plate was inserted, is now occupied by what had been the fifth bright fringe was seen before. Assume the plate have the same thickness t and wavelength of light 480 nm. Then find the value of t in $μ m$ .

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answer is 8.

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Detailed Solution

Five fringes have be shifted.

$Path difference, (μ_{1} - 1) t - (μ_{2} - 1) t = n λ$

$∵ 5 = \frac{(μ_{1} - 1) t - (μ_{2} - 1) t}{λ} = \frac{(μ_{1} - μ_{2}) t}{λ}$

$∴ t = \frac{5 λ}{(μ_{1} - μ_{2})}$

$\Rightarrow t = \frac{5 \times 480 \times 10^{- 9}}{(1.7 - 1.4)} = 8 μ m$

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